A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its position vector is $(-9t^3+9,t^4-16t)$. What is the particle's acceleration vector at $t=2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(-108,48)$ (Choice B) B $\left(9,16\right)$ (Choice C) C $\left(-90,32\right)$ (Choice D) D $\left(-63,-16\right)$
Solution: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's position vector is $(-9t^3+9,t^4-16t)$. We are asked to find the particle's acceleration vector at $t=2$. In other words, we need to find $\vec{a}(2)$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(-9t^3+9),\dfrac{d}{dt}(t^4-16t)\right) \\\\ &=\left(-27t^2,4t^3-16\right) \end{aligned}$ Finding $\vec{a}(t)$ $\begin{aligned} \vec{a}(t)&=\dfrac{d}{dt}\vec{v}(t) \\\\ &=\left(\dfrac{d}{dt}\left(-27t^2\right),\dfrac{d}{dt}(4t^3-16)\right) \\\\ &=\left(-54t,12t^2\right) \end{aligned}$ Finding $\vec{a}(2)$ $\begin{aligned} \vec{a}({2})&=\left(-54({2}),12({2})^2\right) \\\\ &=\left(-108,48\right) \end{aligned}$ In conclusion, the particle's acceleration vector at $t=2$ is $(-108,48)$.